Introduction:

Specifically, only one kind of indicator was utilized for each of the cases and only one reactant was being quantitatively analyzed. Even in the last chapter where an indirect process of analysis was conducted that is Back titration, we utilized only one indicator since the reactants dictated the requirement for the employ of one indicator.

Though, there are cases where more than one indicator is needed for analyzing our model. We shall look into one of such reactions in this chapter as an instance of another technique available to you in volumetric analysis. This technique becomes useful whenever you have a mixture of substances making up a solution.

For example, if we have a solution containing a mixture of NaOH and Na₂CO₃ , that we wish to analyse, the utilize of one indicator to indicate the end of the reaction would be misleading since although an end point was indicated, the 2 substances couldn't have each reacted through the acid completely. To buttress this point, if HCl was titrated by this solution, the reactants that will take place in the same solution are:

1                NaOH + HCl         →     NaCl + H₂O

11 a          Na₂CO₃   + HCl     →     NaHCO₃ + HCl

11 b        NaNCO₃   + HCl        →     NaCl +H₂O + CO₂

Assuming phenolphthalein is utilized as indicator, and then the pink colour of the indicator is discharged whenever reactions 1 and 11a are complete. This is actually the 1^st step of the titration. At this point, all the NaOH is neutralized together by half of the Na₂CO₃ the other half of Na₂CO₃ will transform to NaHCO.

If methyl orange is now inserted and an additional quantity of acid is added, the amount of acid needed will be that necessary to complete reaction 11b. At this stage, the other half of the carbonate is neutralized. Since one mole of NaHCO₃ is formal from one mole of Na₂CO₃ and therefore the quantity of HCl needed for reaction 11 a and 11 b will be similar.

Suppose the volume of the acid required reaching the end-pointing as indicated via the phenolphthalein is a cm³, and the volume of acid reacting with Na₂CO₃ is 2b then the volume of acid reacting by

NaOH is

                   (a + b) - 2bcm³   = (a - b) cm³   

Where:

b cm³  = burette reading at methyl orange end point minus burette reading at phenolphthalein and a cm³   = burette reading at phenolphthalein end point.

Experiment:

Determination of the amount of NaOH and Na₂CO₃ in a specified mixture.

Requirements:

'A' is a standard solution of HCl of known molarity. 'B' is a solution of 2 bases containing unknown quantities of NaOH and Na₂CO₃.

Additional Materials

a) One Burette

b) Phenolphthalein and Methyl orange indicators

c) Unknown concentration of NaOH and Na₂CO₃ mixture.

Procedure:

1^st step: Titration of mixture of bases against acid using Phenolphthalein.

1. Rinse the burette twice by solution 'A' containing HCl and fill it through similar acid. Note the initial burette reading.

2. Rinse the pipette through solution '13' containing a mixture of NaOH and Na₂CO₃ and take 25cm³ portion of this solution into a 250cm³ conical flask and add 2-3 drops of Phenolphthalein.

3. Add acid from the burette until pink colour disappears. Note the burette reading of the Phenolphthalein.

2^nd Step: Continuation of titration through similar solutions, with no adding acid into the burette or base into the conical flask, using methyl orange (1 drop) as indicator.

4. Add one drop of methyl orange indicator to the colourless solution attained at the Phenolphthalein end point and continue adding HCl from the burette until one drop of acid gives colour. At this point the remaining NaHCO₃ is neutralized. Note the burette reading again. This is the methyl orange end -point.

5. Now pipette another 25cm³ portion of solution 'B' (base) again into a clean conical flask, fill the burette through HCl solution and start the titration watchfully, 1^st using Phenolphthalein. As in step one and then methyl orange as in step two.

6. Repeat the titration until 3 constant consequences are attained.

Problem:

From your average titre values, determine

a) The concentration in moles per dm³ of solution NaOH that has reacted

b) The concentration in moles per dm³ of solution Na₂CO₃ that has reacted.

c) The amount in grams of the NaOH in the mixture

d) The amount in mass of Na₂CO₃ in the mixture

e) The percent composition of NaOH in the mixture

f) The percent composition of Na₂CO₃ in the mixture Na = 23, 0 = 16. C = 12, H =1).

Results:

Treatment of Results:

Calculations:

a) For the amount in mole,/ dm³ of NaOH:

NaOH+ HCl     →   NaCl+ H2O

1^st. We find the amount in moles of HCl = Molar cone. X Volume /1000

= Molar conc. x (a - b ) cm³  /1000

= C moles

From the stoichiciometry. I mole HCl = 1 mole NaOH

Therefore Amount in moles of C moles HCl = C moles of NaOH

This C mole is enclosed in the 25cm³ of NaOH utilized in the titration.

Therefore 1000cm³ will enclose

= C moles x 1000 /25

= 40 C moles/ dm³   

b) For the amount in moles / dm³ of Na₂CO₃:

Na₂CO₃ + 2HCl   →    2NaCl + H₂O

First, we find the amount in moles of HCI

= molar conc. x volume / 1000

= molar conc. x (2b) cm³ /1000

= D moles

From the stoichiciometry, 2 mole HCl = 1 mole NaOH

Therefore Amount in moles of D moles HCl = 1 /2 D moles of Na₂CO₃

Hence 1/2 D mole is contained in the 25cm³ of Na₂CO₃   used in the titration.

Therefore 1000cm³ will contain = 1/2 D moles x 1000 / 25

                                                  = 20 D moles/ dm³

c) Amount in g/dm³   of NaOH

Concentration in g/dm³ = molar mass x molar /conc. of NaOH

                                     = 40 x 40C

                                    = 1600C g/dm³   

d) Amount in g/dm³ of Na₂CO₃=molar mass x molar /conc. of Na₂CO₃

                                          = 106 x 20D

                                          = 200D/gdm³  

e) Percentage of the NaOH in the mixture.

                 % of NaOH = Mass of NaOH perdm³ x 100 / the combined mass of the two bases.

f) Percentage of the NA₂CO₃ in the mixture

              % of Na₂CO₃   = Mass of N₂CO₃ perdm³   x 100 / The combined mass of the two bases

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